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The solubility of saturated solution of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $s$ mol $\mathrm{L}^{-1}$. What is its solubility product?
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Verified Answer
The correct answer is:
$4 s^3$
$\mathrm{Ag}_2 \mathrm{CrO}_4$ ionises completely in the solution as
$\underset{s \mathrm{~mol} / \mathrm{L}}{\mathrm{Ag}_2 \mathrm{CrO}_4 \longrightarrow} \underset{2 s \mathrm{~mol} / \mathrm{L}}{2 \mathrm{Ag}^{+}}+\underset{s \mathrm{~mol} / \mathrm{L}}{\mathrm{CrO}_4^{2-}}$
Hence, the solubility product,
$\begin{aligned}
K_{\text {sp }} \text { of } \mathrm{Ag}_2 \mathrm{CrO}_4 & =\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CrO}_4^{2-}\right] \\
& =(2 s)^2(s) \\
& =4 s^3
\end{aligned}$
$\underset{s \mathrm{~mol} / \mathrm{L}}{\mathrm{Ag}_2 \mathrm{CrO}_4 \longrightarrow} \underset{2 s \mathrm{~mol} / \mathrm{L}}{2 \mathrm{Ag}^{+}}+\underset{s \mathrm{~mol} / \mathrm{L}}{\mathrm{CrO}_4^{2-}}$
Hence, the solubility product,
$\begin{aligned}
K_{\text {sp }} \text { of } \mathrm{Ag}_2 \mathrm{CrO}_4 & =\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CrO}_4^{2-}\right] \\
& =(2 s)^2(s) \\
& =4 s^3
\end{aligned}$
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